Computational Genomics Lecture 1, Tuesday April 1, 2003.

Computational Genomics

Lecture 1, Tuesday April 1, 2003


Biology in One Slide


High Throughput Biology

1. DNA Sequencing

…ACGTGACTGAGGACCGTGCGACTGAGACTGACTGGGTCTAGCTAGACTACGTTTTATATATATATACGTCGTCGTACTGATGACTAGATTACAGACTGATTTAGATACCTGACTGATTTTAAAAAAATATT…



2. Sequencing of expressed genes(EST sequencing)

mRNA sequence

protein sequence



3. Gene Expression: Microarrays



4. Gene Regulation: CH.IP.


The goals of genomics

• Study organisms at the DNA level

– Identify “parts” (genes, etc)– Figure out “connections” between

“parts”

• Study evolution at the DNA level

– Compare organisms– Uncover evolutionary history


The role of CS in Biology

Essential– DNA sequencing and assembly– Microarray analysis– Protein 3D reconstruction

Complementary– Gene finding, genome annotation– Protein fold prediction– Phylogeny, comparative

genomics


Syllabus

• Tools

– Alignment algorithms– Hidden Markov models– Statistical algorithms

• Applications

– DNA sequencing and assembly– Sequence analysis (comparison, annotation)– Microarray analysis– Evolutionary analysis


Course responsibilities

• Homeworks [80%]– 4 challenging problem sets, 4-5 problems/pset– Collaboration allowed– 5 late days total– Televised students required to do 75%

• Final [20%]– Takehome, 1 day– Collaboration not allowed– Easy!

• Scribing– “Mandatory”– Grade replaces lowest 2 problems– Due one week after the lecture


Reading material

• Books– “Biological sequence analysis” by

Durbin, Eddy, Krogh, Mitchinson

• Chapters 1-4, 6, (7-8), (9-10)

– “Algorithms on strings, trees, and sequences” by Gusfield

• Chapters (5-7), 11-12, (13), 14, (17)

• Papers• Lecture notes

Topic 1. Sequence Alignment


Complete genomes


Evolution


Evolution at the DNA level

…ACGGTGCAGTCACCA…

…ACGTTGCAGTCCACCA…

C

SEQUENCE EDITS REARRANGEMENTS


Evolutionary Rates

OK

OK

OK

X

X

Still OK?

next generation


Sequence conservation implies function

Interleukin region in human and mouse


Sequence Alignment

-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---TAG-CTATCAC--GACCGC--GGTCGATTTGCCCGAC

DefinitionGiven two strings x = x1x2...xM, y

= y1y2…yN,

an alignment is an assignment of gaps to positions

0,…, N in x, and 0,…, N in y, so as to line up each letter in one sequence with either a letter, or a gap

in the other sequence

AGGCTATCACCTGACCTCCAGGCCGATGCCCTAGCTATCACGACCGCGGTCGATTTGCCCGAC


What is a good alignment?

Alignment: The “best” way to match the letters of one sequence with those of the other

How do we define “best”?

Alignment:A hypothesis that the two sequences come from a common ancestor through sequence edits

Parsimonious explanation:Find the minimum number of edits that transform one sequence into the other


Scoring Function

• Sequence edits:AGGCCTC

– Mutations AGGACTC

– InsertionsAGGGCCTC

– DeletionsAGG.CTC

Scoring Function:Match: +mMismatch: -sGap: -d

Score F = (# matches) m - (# mismatches) s – (#gaps) d


How do we compute the best alignment?

AGTGCCCTGGAACCCTGACGGTGGGTCACAAAACTTCTGGA

AGTGACCTGGGAAGACCCTGACCCTGGGTCACAAAACTC

Too many possible alignments:

O( 2M+N)


Alignment is additive

Observation:The score of aligning x1……xM

y1……yN

is additive

Say that x1…xi xi+1…xM

aligns to y1…yj yj+1…yN

The two scores add up:

F(x[1:M], y[1:N]) = F(x[1:i], y[1:j]) + F(x[i+1:M], y[j+1:N])


Dynamic Programming

• We will now describe a dynamic programming algorithm

Suppose we wish to alignx1……xM

y1……yN

Let F(i,j) = optimal score of aligning

x1……xi

y1……yj


Dynamic Programming (cont’d)

Notice three possible cases:

1. xi aligns to yj

x1……xi-1 xi

y1……yj-1 yj

2. xi aligns to a gap

x1……xi-1 xi

y1……yj -

3. yj aligns to a gap

x1……xi -

y1……yj-1 yj

m, if xi = yj

F(i,j) = F(i-1, j-1) + -s, if not

F(i,j) = F(i-1, j) - d

F(i,j) = F(i, j-1) - d


Dynamic Programming (cont’d)

• How do we know which case is correct?

Inductive assumption:F(i, j-1), F(i-1, j), F(i-1, j-1) are optimal

Then,F(i-1, j-1) + s(xi, yj)

F(i, j) = max F(i-1, j) – dF( i, j-1) – d

Where s(xi, yj) = m, if xi = yj; -s, if not


Example

x = AGTA m = 1y = ATA s = -1

d = -1

A G T A

0 -1 -2 -3 -4

A -1 1 0 -1 -2

T -2 0 0 1 0

A -3 -1 -1 0 2

F(i,j) i = 0 1 2 3 4

j = 0

1

2

3

Optimal Alignment:

F(4,3) = 2

AGTAA - TA


The Needleman-Wunsch Matrix

x1 ……………………………… xM

y1 …

……

……

……

……

……

…

yN

Every nondecreasing path

from (0,0) to (M, N)

corresponds to an alignment of the two sequences

Can think of it as adivide-and-conquer algorithm


The Needleman-Wunsch Algorithm

1. Initialization.a. F(0, 0) = 0b. F(0, j) = - j dc. F(i, 0) = - i d

2. Main Iteration. Filling-in partial alignmentsa. For each i = 1……M

For each j = 1……N F(i-1,j) – d [case

1]F(i, j) = max F(i, j-1) – d [case

2] F(i-1, j-1) + s(xi, yj)

[case 3]

UP, if [case 1]Ptr(i,j) = LEFT if [case 2]

DIAG if [case 3]

3. Termination. F(M, N) is the optimal score, andfrom Ptr(M, N) can trace back optimal alignment


Performance

• Time:O(NM)

• Space:O(NM)

• Later we will cover more efficient methods


A variant of the basic algorithm:

• Maybe it is OK to have an unlimited # of gaps in the beginning and end:

----------CTATCACCTGACCTCCAGGCCGATGCCCCTTCCGGCGCGAGTTCATCTATCAC--GACCGC--GGTCG--------------

• Then, we don’t want to penalize gaps in the ends


Different types of overlaps


The Overlap Detection variant

Changes:

1. InitializationFor all i, j,

F(i, 0) = 0F(0, j) = 0

2. Termination maxi

F(i, N)FOPT = max

maxj F(M, j)

x1 ……………………………… xM

y1 …

……

……

……

……

……

…

yN


Next Lecture

• Local alignment

• More elaborate scoring function

• Memory-efficient algorithms

Reading:Durbin, Chapter 2

Gusfield, Chapter 11

Computational Genomics Lecture 1, Tuesday April 1, 2003.

Documents

Transcript of Computational Genomics Lecture 1, Tuesday April 1, 2003.